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[tex]\sf\lim\limits_{x \to 0}\frac{8x\cos(5x)}{{x}^{2} + 2x + \sin(4x)}=...[/tex] - Ngasal, gk

Berikut ini adalah pertanyaan dari unknown pada mata pelajaran Matematika untuk jenjang Sekolah Menengah Pertama

\sf\lim\limits_{x \to 0}\frac{8x\cos(5x)}{{x}^{2} + 2x + \sin(4x)}=...-
Ngasal, gk ada cara => report
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Jawaban dan Penjelasan

Berikut ini adalah pilihan jawaban terbaik dari pertanyaan diatas.

\sf\lim\limits_{x \to 0}\frac{8x\cos(5x)}{{x}^{2} + 2x + \sin(4x)}=\huge \boxed {\sf0}

Pengerjaan Digambar

• Sifat Turunan Aljabar

y = axⁿ → y' = a.n.x^n-1

Contoh:

y = 4x²

y' = 4.2.x^2-1

y' = 8x^1

y' = 8x

_____.

• Sifat Turunan Trigonometri

y = cos x → y' = sin x

y = cos ax → y' = a sin ax

y = sin x → y' = cos x

y = sin ax → y' = a cos ax

___.

Tambahan

sin a/cos a → tan a

tan 0 = 0

Jawab:4/3Penjelasan dengan langkah-langkah:[tex]\large\text{$\begin{aligned}&\lim\limits_{x\to0}\left(\frac{8x\cos(5x)}{{x}^{2}+2x+\sin(4x)}\right)\\&{=\:}\lim\limits_{x\to0}\left(8\cdot\frac{x\cos(5x)}{{x}^{2}+2x+\sin(4x)}\right)\\&{=\:}8\cdot\lim\limits_{x\to0}\left(\frac{x\cos(5x)}{{x}^{2}+2x+\sin(4x)}\right)\\&{=\:}8\cdot\lim\limits_{x\to0}\left(\frac{\cancel{x}\big(\cos(5x)\big)}{\cancel{x}\left(x+2+\frac{\sin(4x)}{x}\right)}\right)\\&{=\:}8\cdot\lim\limits_{x\to0}\left(\frac{\cos(5x)}{x+2+\frac{\sin(4x)}{x}}\right)\end{aligned}$}[/tex][tex]\large\text{$\begin{aligned}&{=\:}8\cdot\frac{\lim\limits_{x\to0}{\:\cos(5x)}}{\lim\limits_{x\to0}\:\left(x+2+\frac{\sin(4x)}{x}\right)}\\&{=\:}8\cdot\frac{\lim\limits_{x\to0}{\:\cos(5x)}}{\lim\limits_{x\to0}\:(x+2)+\lim\limits_{x\to0}\left(\frac{\sin(4x)}{x}\right)}\end{aligned}$}[/tex][tex]\large\text{$\begin{aligned}&\quad\normalsize\textsf{..... aturan L'H\^opital}\\&\quad\left|\normalsize\text{$\begin{aligned}\lim\limits_{x\to0}\left(\frac{\sin(4x)}{x}\right)&=\lim\limits_{x\to0}\left(\frac{\frac{d}{dx}\sin(4x)}{\frac{d}{dx}x}\right)\\&=\lim\limits_{x\to0}\left(\frac{4\cos(4x)}{1}\right)\\&=\lim\limits_{x\to0}(4\cos(4x))\\\end{aligned}$}\right.\\\\&{=\:}8\cdot\frac{\lim\limits_{x\to0}{\:\cos(5x)}}{\lim\limits_{x\to0}\:(x+2)+\lim\limits_{x\to0}(4\cos(4x))}\\\end{aligned}$}[/tex][tex]\large\text{$\begin{aligned}&\quad\normalsize\textsf{..... substitusi $x$}\\&{=\:}8\cdot\frac{\cos(0)}{(0+2)+4\cos(0)}\\&{=\:}8\cdot\frac{1}{2+4\cdot1}=\frac{8}{6}\\&{=\:}\boxed{\ \bf\frac{4}{3}\ }\end{aligned}$}[/tex]Jawab:4/3Penjelasan dengan langkah-langkah:[tex]\large\text{$\begin{aligned}&\lim\limits_{x\to0}\left(\frac{8x\cos(5x)}{{x}^{2}+2x+\sin(4x)}\right)\\&{=\:}\lim\limits_{x\to0}\left(8\cdot\frac{x\cos(5x)}{{x}^{2}+2x+\sin(4x)}\right)\\&{=\:}8\cdot\lim\limits_{x\to0}\left(\frac{x\cos(5x)}{{x}^{2}+2x+\sin(4x)}\right)\\&{=\:}8\cdot\lim\limits_{x\to0}\left(\frac{\cancel{x}\big(\cos(5x)\big)}{\cancel{x}\left(x+2+\frac{\sin(4x)}{x}\right)}\right)\\&{=\:}8\cdot\lim\limits_{x\to0}\left(\frac{\cos(5x)}{x+2+\frac{\sin(4x)}{x}}\right)\end{aligned}$}[/tex][tex]\large\text{$\begin{aligned}&{=\:}8\cdot\frac{\lim\limits_{x\to0}{\:\cos(5x)}}{\lim\limits_{x\to0}\:\left(x+2+\frac{\sin(4x)}{x}\right)}\\&{=\:}8\cdot\frac{\lim\limits_{x\to0}{\:\cos(5x)}}{\lim\limits_{x\to0}\:(x+2)+\lim\limits_{x\to0}\left(\frac{\sin(4x)}{x}\right)}\end{aligned}$}[/tex][tex]\large\text{$\begin{aligned}&\quad\normalsize\textsf{..... aturan L'H\^opital}\\&\quad\left|\normalsize\text{$\begin{aligned}\lim\limits_{x\to0}\left(\frac{\sin(4x)}{x}\right)&=\lim\limits_{x\to0}\left(\frac{\frac{d}{dx}\sin(4x)}{\frac{d}{dx}x}\right)\\&=\lim\limits_{x\to0}\left(\frac{4\cos(4x)}{1}\right)\\&=\lim\limits_{x\to0}(4\cos(4x))\\\end{aligned}$}\right.\\\\&{=\:}8\cdot\frac{\lim\limits_{x\to0}{\:\cos(5x)}}{\lim\limits_{x\to0}\:(x+2)+\lim\limits_{x\to0}(4\cos(4x))}\\\end{aligned}$}[/tex][tex]\large\text{$\begin{aligned}&\quad\normalsize\textsf{..... substitusi $x$}\\&{=\:}8\cdot\frac{\cos(0)}{(0+2)+4\cos(0)}\\&{=\:}8\cdot\frac{1}{2+4\cdot1}=\frac{8}{6}\\&{=\:}\boxed{\ \bf\frac{4}{3}\ }\end{aligned}$}[/tex]Jawab:4/3Penjelasan dengan langkah-langkah:[tex]\large\text{$\begin{aligned}&\lim\limits_{x\to0}\left(\frac{8x\cos(5x)}{{x}^{2}+2x+\sin(4x)}\right)\\&{=\:}\lim\limits_{x\to0}\left(8\cdot\frac{x\cos(5x)}{{x}^{2}+2x+\sin(4x)}\right)\\&{=\:}8\cdot\lim\limits_{x\to0}\left(\frac{x\cos(5x)}{{x}^{2}+2x+\sin(4x)}\right)\\&{=\:}8\cdot\lim\limits_{x\to0}\left(\frac{\cancel{x}\big(\cos(5x)\big)}{\cancel{x}\left(x+2+\frac{\sin(4x)}{x}\right)}\right)\\&{=\:}8\cdot\lim\limits_{x\to0}\left(\frac{\cos(5x)}{x+2+\frac{\sin(4x)}{x}}\right)\end{aligned}$}[/tex][tex]\large\text{$\begin{aligned}&{=\:}8\cdot\frac{\lim\limits_{x\to0}{\:\cos(5x)}}{\lim\limits_{x\to0}\:\left(x+2+\frac{\sin(4x)}{x}\right)}\\&{=\:}8\cdot\frac{\lim\limits_{x\to0}{\:\cos(5x)}}{\lim\limits_{x\to0}\:(x+2)+\lim\limits_{x\to0}\left(\frac{\sin(4x)}{x}\right)}\end{aligned}$}[/tex][tex]\large\text{$\begin{aligned}&\quad\normalsize\textsf{..... aturan L'H\^opital}\\&\quad\left|\normalsize\text{$\begin{aligned}\lim\limits_{x\to0}\left(\frac{\sin(4x)}{x}\right)&=\lim\limits_{x\to0}\left(\frac{\frac{d}{dx}\sin(4x)}{\frac{d}{dx}x}\right)\\&=\lim\limits_{x\to0}\left(\frac{4\cos(4x)}{1}\right)\\&=\lim\limits_{x\to0}(4\cos(4x))\\\end{aligned}$}\right.\\\\&{=\:}8\cdot\frac{\lim\limits_{x\to0}{\:\cos(5x)}}{\lim\limits_{x\to0}\:(x+2)+\lim\limits_{x\to0}(4\cos(4x))}\\\end{aligned}$}[/tex][tex]\large\text{$\begin{aligned}&\quad\normalsize\textsf{..... substitusi $x$}\\&{=\:}8\cdot\frac{\cos(0)}{(0+2)+4\cos(0)}\\&{=\:}8\cdot\frac{1}{2+4\cdot1}=\frac{8}{6}\\&{=\:}\boxed{\ \bf\frac{4}{3}\ }\end{aligned}$}[/tex]

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Last Update: Wed, 20 Apr 22
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